# Cyclic cubic numbers as rational linear combinations of roots of unity

In the written version of a talk Barry Mazur gave to Friends of the Harvard Mathematics Department on May 5, 2009, there is an interesting question in Footnote 5 (page 8).

He recalls how Gauss wrote $\sqrt p$ (where $p$ is an odd prime) as an explicit rational linear combination of roots of unity (using Gauss sums) and says that he doesn't know any such explicit expression for the roots $\alpha$ of an irreducible cubic polynomial $T^3+bT+c\in\mathbf{Q}[T]$ whose discriminant is a square (so that $\mathbf{Q}(\alpha)$ is a cyclic extension of $\mathbf{Q}$, and hence contained in $\mathbf{Q}(\zeta)$ for some root of unity $\zeta$).

Question. Does anyone know such an explicit expression for the roots of irreducible cubic polynomials whose discriminant is a square ?

• Chapter 3 of H. Davenport, Multiplicative Number Theory, Third Edition, has some relevant material on this, I think. Jan 4, 2012 at 9:54

I guess Mazur's remark just means that in the expression of the cubic polynomial whose roots generate the cubic subfield of the $p$-th roots of unity, there are numbers $L$ and $M$ with $L^2 + 27M^2 = p$. It reflects the fact from Kummer theory that for understanding cyclic cubic extensions you have to adjoin the cube roots of unity, whereas for quadratic extensions the necessary square roots of unity are already there.
For the simplest cubic $x^3-ax^2-(a+3)x-1$ which is cyclic and real with discriminant $p^2$ where $p=a^2+3a+9$, and when $p$ is prime, the roots $\theta_j,j=0,1,2$ are translates of the Gauss's cubic periods $\eta_j$. Explicitly $\theta_j=\eta_j+(L-1)/6$, where $4p=L^2+27$ and $\eta_j$ are the Gauss's periods. Since $\sum_{j=0}^2 \eta_j=-1$, one can certainly replace $(L-1)/6$ by linear combinations of $\eta_j$. This relation appeared (and also the quartic and sextic case) in E. Lehmer, "Connections between Gaussian periods and cyclotomic units", Maths Comp. Vol 50, No. 182 (1988) 535-541.
• At least for this case, the polynomials are also "nice" in the sense I have just defined in the MO thread mathoverflow.net/questions/86401/…. E.g., for $a=1$ we can write the roots as $2(\cos\frac{k\pi}{13}+\cos\frac{5k\pi}{13}),\ k=1,3,9$. for $a=2$ as $1-2 (\cos\frac{k\pi}{19}+\cos\frac{7k\pi}{19}+\cos\frac{11k\pi}{19}),\ k=1,3,9$, for $a=3$ as $1-\sqrt{12}\sin\frac{k\pi}{9},\ k=1,2,4$, etc. Jan 22, 2012 at 21:57
• Yes, since the maxmimal real subfield of the field generated by $e^{2\pi i k/n)$ is the one generated by $\cos ( 2 \pi k/n)$. Jan 22, 2012 at 22:22